Consolidation settlement worksheet with answers

Free printable Consolidation settlement worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
Δσ = 60 kPasandclayCc​ = 0.4e0​ = 1.05σ′0​ = 65 kPa8.5 mwater table

An 8.5 m thick layer of normally consolidated clay (Cc = 0.4, e₀ = 1.05) has an effective vertical stress of 65 kPa at its mid-depth. A wide fill raises the vertical stress in the clay by 60 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

H = 8.5 mCc = 0.4e0 = 1.05σ′0 = 65 kPaΔσ = 60 kPa
S = mm
2.Easy
Δσ = 140 kPasandγ = 18.5 kN/m³2 mclayγsat​ = 18.5 kN/m³Cc​ = 0.45, e0​ = 0.85 mwater tableX

A 5 m thick normally consolidated clay layer (γsat = 18.5 kN/m³, Cc = 0.45, e₀ = 0.8) lies below 2 m of sand (γ = 18.5 kN/m³). The water table is at the top of the clay; use γw = 9.81 kN/m³. A wide fill raises the vertical stress throughout the clay by 140 kPa. Treat the layer as one sublayer with the stresses at its mid-depth. Find the primary consolidation settlement S in mm.

sand: Hs = 2 m, γ = 18.5 kN/m³clay: H = 5 m, γsat = 18.5 kN/m³γw = 9.81 kN/m³Cc = 0.45e0 = 0.8Δσ = 140 kPa
S = mm
3.Medium
Δσ = 60 kPasandclayCc​ = 0.4, Cr​ = 0.02e0​ = 0.95σ′0​ = 100 kPaOCR = 25.5 mwater table

A 5.5 m thick overconsolidated clay layer has Cc = 0.4, Cr = 0.02 and e₀ = 0.95. At its mid-depth the effective vertical stress is 100 kPa and the overconsolidation ratio is OCR = 2. A new building adds 60 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 5.5 mCc = 0.4Cr = 0.02e0 = 0.95σ′0 = 100 kPaOCR = 2Δσ = 60 kPa
S = mm
4.Medium
Δσ = 180 kPasandclayCc​ = 0.25, Cr​ = 0.04e0​ = 0.7σ′0​ = 100 kPaOCR = 1.56 mwater table

A 6 m thick overconsolidated clay layer has Cc = 0.25, Cr = 0.04 and e₀ = 0.7. At its mid-depth the effective vertical stress is 100 kPa and the overconsolidation ratio is OCR = 1.5. A new building adds 180 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.

H = 6 mCc = 0.25Cr = 0.04e0 = 0.7σ′0 = 100 kPaOCR = 1.5Δσ = 180 kPa
S = mm
5.Hard
sandclayH = 6 mcv​ = 1.2 m²/year6 mimpervious rockwater table

A clay layer 6 m thick has a coefficient of consolidation cv = 1.2 m²/year. It lies on impervious rock, with sand above it, so water drains out only at the top. For 90% average consolidation the time factor is Tv = 0.848. How many years t does the layer take to reach 90% consolidation?

H = 6 mcv = 1.2 m²/yearU = 90%Tv = 0.848single drainage
t = years
6.Hard
sandclayH = 5 m5 mimpervious rockwater table

In a laboratory oedometer test, a clay specimen 20 mm thick, drained at the top and bottom, reaches 50% consolidation in 12 min. In the field the same clay forms a layer 5 m thick. It lies on impervious rock, with sand above it, so water drains out only at the top. How many days t does the field layer take to reach 50% consolidation?

hlab = 20 mmtlab = 12 minH = 5 mfield: single drainage
t = days

Answer key

  1. 1. S = 471 mm
  2. 2. S = 661.8 mm
  3. 3. S = 11.51 mm
  4. 4. S = 264 mm
  5. 5. t = 25.44 years
  6. 6. t = 2083 days

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